1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Work required to increase the radius of a sphere by dr

  1. Oct 2, 2013 #1
    1. The problem statement, all variables and given/known data
    To assemble a uniformly charged sphere, assemble it like a snowball, layer by layer, each time bringing in an infinitesimal charge dq from far away and smearing it uniformly over the surface, thereby increasing the radius. How much work dW does it take to build up the radius by an amount dr? Integrate this to find the work necessary to create the entire sphere of radius R and total charge q.


    2. Relevant equations

    W = 1/2∫ρVdτ

    3. The attempt at a solution

    dW = k*q*(dq/R)

    This is the work needed to add a charge dq from infinity to the outer shell of the sphere

    it will then increase it by dr based on the charge density added to the sphere.

    so dq = 4πr^2*ρ*q*dr

    making dW = (k*q*4πr^2*ρ*dr)/R

    So W = k*q*4πr^2*∫ρ r^2 dr dτ

    I am confused if i have set this up right and on what interval i need to integrate.
     
    Last edited: Oct 3, 2013
  2. jcsd
  3. Oct 3, 2013 #2

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    For your final expression for dW, write out in words what it means.
    It is the work needed to...
     
  4. Oct 3, 2013 #3
    dW is needed to take a charge from infinitely far away and distribute it even across the surface of the sphere thus increasing the radius by an infinitesimally small amount.

    Thus making W to make the sphere be the the sum of dW from 0 to R which answers my integration from where problem. Thank you for that.

    I am still confused on if i have set up the problem right though. So would it be correct to just integrate it so that

    W = k*q*4πr^2*ρ* ∫ r^2 dr from 0 to R?
     
  5. Oct 3, 2013 #4

    rude man

    User Avatar
    Homework Helper
    Gold Member

    dW = k q1 q2 /r in general
    Here, q1 is the charge q as a function of r. What is that? Hint: charge = charge density x volume.
    And here q2 = dq.
    So express dW as a function of r and dq: dW = k q(r) dq/r
    Then, take dq = f(r) dr and substitute so that dW does not have q in it. You have eliminated q and dq in your equation for dW. Then integrate from r = 0 to r = R.

    And whatever power of r you wind up with, it all has to go INSIDE the integral sign!
     
    Last edited: Oct 3, 2013
  6. Oct 3, 2013 #5

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    I'm thinking - doesn't the amount of work you have to do also depend on how much charge has already been moved? Or did you take account of that later?

    ... dq would be the amount of charge in a spherical shell thickness dr?
    i.e. the surface area of the underlying volume radius r times the thickness. That how you did it?

    You've already seen that writing down your reasoning is good for sorting out your thoughts later. It is also good for getting those elusive final marks on long-answer questions.
     
  7. Oct 3, 2013 #6
    Thank you, i was able to the problem out.
     
  8. Oct 3, 2013 #7

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Well done - want to share for those who google here with the same question?
     
  9. Oct 3, 2013 #8

    rude man

    User Avatar
    Homework Helper
    Gold Member

    Yes, I would like to compare my answers with yours.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Work required to increase the radius of a sphere by dr
  1. Radius of Fermi Sphere (Replies: 1)

Loading...